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Chapter 8: Problem 61

Arrange the following groups of molecules in order of increasing dipole moment: \(\begin{array}{llll}\text { (a) } \mathrm{HCl} & \mathrm{HF} & \mathrm{HI} & \mathrm{HBr}\end{array}\) \(\begin{array}{lll}\text { (b) } \mathrm{PH}_{\mathrm{s}} & \mathrm{NH}_{\mathrm{y}} & \mathrm{AsH}_{3}\end{array}\) \(\begin{array}{lll}\text { (c) } \mathrm{ClF}_{3} & \mathrm{BrF}_{3} & \mathrm{IF}_{\mathrm{s}}\end{array}\) \(\begin{array}{llll}\text { (d) } \mathrm{H}_{2} \mathrm{O} & \mathrm{H}_{2} \mathrm{~S} & \mathrm{H}_{2} \mathrm{Te} & \mathrm{H}_{2} \mathrm{Se}\end{array}\)

Short Answer

Expert verified
(a) HI < HBr < HCl < HF, (b) AsH3 < PH3 < NH3, (c) BrF3 < ClF3 < IF5, (d) H2Te < H2Se < H2S < H2O.

Step by step solution

01

Understanding the Dipole Moment

A dipole moment arises from differences in electronegativity between atoms in a molecule. It measures the separation of positive and negative charges in a bond or molecule. Molecules with greater differences in electronegativity generally have larger dipole moments.
02

Analyzing Group (a) - Halogen Hydrides

In group (a), the molecules are HCl, HF, HI, and HBr. Electronegativity decreases in the order F > Cl > Br > I. Since HF has the largest electronegativity difference, it has the highest dipole moment. Conversely, HI has the smallest, as iodine has the least electronegativity. The order of increasing dipole moment is: HI < HBr < HCl < HF.
03

Analyzing Group (b) - Hydrides of Group 15 Elements

For PH3, NH3, and AsH3, we consider the central atom's electronegativity and the geometry. NH3 is the most polar due to nitrogen's higher electronegativity and its more compact shape contributing to a larger dipole. PH3 and AsH3 have much smaller dipoles, with AsH3 being the smallest due to arsenic's larger size and lower electronegativity. Thus, the order is: AsH3 < PH3 < NH3.
04

Analyzing Group (c) - Fluorides of Halogens

In group (c), ClF3, BrF3, and IF5, the dipole moment depends heavily on molecular shape (T-shaped for ClF3 and BrF3, and square pyramidal for IF5). The larger the central atom, the smaller its electronegativity effect becomes, but the difference is also subtle due to similar structures. Therefore, the order is determined more by geometry rather than direct electronegativity differences: BrF3 < ClF3 < IF5.
05

Analyzing Group (d) - Hydrides of Chalcogens

In H2O, H2S, H2Te, and H2Se, we evaluate based on electronegativity. Oxygen is the most electronegative, leading to the largest dipole in H2O. While H2S is less electronegative than H2Se, the difference is compensated by its smaller size, making H2S' dipole moment smaller. H2Te, with the least electronegativity and larger size, has the smallest dipole. Order is: H2Te < H2Se < H2S < H2O.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Electronegativity
Electronegativity is a measure of an atom's ability to attract and hold electrons within a chemical bond. Each element has a specific electronegativity value, which influences how atoms bond and the distribution of electron density in molecules.
  • Fluorine is the most electronegative element, which means it pulls electrons towards itself more than any other atom.
  • Electronegativity trends within the periodic table generally increase across a period (from left to right) and decrease down a group (from top to bottom).
In molecules, differences in electronegativity between bonded atoms can lead to a separation of charges, creating a dipole moment. This is why electronegativity is crucial when predicting the behavior of polar and nonpolar molecules.
Molecular Geometry
Molecular geometry refers to the three-dimensional arrangement of atoms within a molecule. It plays a crucial role in determining physical properties and chemical reactivity.
Different molecular geometries can lead to differences in dipole moments, depending on how atom positions and electron pairs are arranged.
  • Linear, bent, tetrahedral, trigonal planar, and octahedral are some basic molecular geometries.
  • The geometry is influenced by the electron pair repulsion and bonding requirements of the central atom.
The molecular geometry can either enhance or cancel out dipole moments, which is important when assessing a molecule's polarity.
Polar Molecules
Polar molecules have a net dipole moment because they contain bonds with uneven distribution of electrons, due to differences in electronegativity. Essentially, one end of the molecule is slightly positive, while the other is slightly negative.
  • Water (H鈧侽) is a classic example of a polar molecule, because of its bent molecular structure and high electronegativity of oxygen.
  • To determine if a molecule is polar, we assess both its electronegativity differences and its geometrical arrangement.
Polar molecules interact through dipole-dipole interactions, affecting their boiling and melting points, solubility, and overall reactivity. They are generally much different chemically and physically compared to nonpolar molecules, which do not have a permanent dipole moment.
Halogen Hydrides
Halogen hydrides (HX) are binary compounds where a halogen (X) is bonded to a hydrogen atom. These hydrides include HF, HCl, HBr, and HI, each having unique properties arising from differences in their electronegativity and size.
  • Fluorine, being the most electronegative halogen, forms a very polar HF molecule with a strong dipole moment.
  • In contrast, iodine's low electronegativity results in HI having a much smaller dipole moment.
The acidity and boiling points of these hydrides are influenced by both bond strength and the polarity of the bond, making them an interesting group of molecules to study in terms of chemical behavior.
Group 15 Hydrides
Group 15 hydrides include ammonia (NH鈧�), phosphine (PH鈧�), and arsine (AsH鈧�), which are collectively known as pnictogen hydrides. The central atom in these molecules is from Group 15 of the periodic table and exhibits distinct properties based on its size and electronegativity.
  • Ammonia, with nitrogen as the central atom, has a pronounced dipole moment due to nitrogen's high electronegativity.
  • Phosphine and arsine have much weaker dipole moments, arising from their larger, less electronegative central atoms.
Thus, these molecules demonstrate how central atom characteristics, such as size and electronegativity, contribute to the polarity and other properties of the hydrides.
Fluorides of Halogens
Fluorides of halogens are compounds such as ClF鈧�, BrF鈧�, and IF鈧�, where fluorine atoms are bonded to other halogens. These molecules have varying molecular geometries that cause differences in their polarity and dipole moments.
  • For instance, ClF鈧� has a T-shaped molecular geometry, affecting its dipole and properties.
  • IF鈧� exhibits a square pyramidal shape, contributing to a higher dipole moment than its counterparts.
The interplay between molecular shape and the electronegativity of fluorine is pivotal in determining the chemical and physical behaviors of these compounds, making them an interesting topic in advanced chemistry studies.
Hydrides of Chalcogens
Hydrides of chalcogens include compounds like H鈧侽, H鈧係, H鈧係e, and H鈧俆e, where chalcogen elements form bonds with hydrogen. Oxygen, sulfur, selenium, and tellurium are the central atoms that bond with hydrogens, each offering different properties based on their electronegativity and atomic size.
  • Water (H鈧侽) is extremely polar due to oxygen's high electronegativity creating a large dipole moment.
  • In contrast, the larger and less electronegative tellurium in H鈧俆e results in a weaker dipole moment.
The unique characteristics of these molecules, especially water's well-known polarity, make chalcogen hydrides a significant area of study for understanding solubility, chemical reactivity, and other physical properties.

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Most popular questions from this chapter

Predict the shape and bond angles of each of the following species: (a) \(\mathrm{XeOF}_{4}\) (b) IOF, (c) \(\mathrm{PO}_{2} \mathrm{~F}_{2}^{-}\) (d) \(\mathrm{PO}_{3} \mathrm{~F}^{2-}\)

For each of the following molecules, predict the shape and bond angles. Indicate which ones are polar. (a) \(\mathrm{TeF}_{6}\) (b) \(\mathrm{ClF}_{5}\) (c) \(\mathrm{SiCl}_{4}\) (d) \(\mathrm{SeCl}_{2}\)

Give the molecular class, shape, and bond angles \(\left(90^{\circ}, 109.5^{\circ},<120^{\circ}\right.\), etc.) for each of the following halides: (a) \(\mathrm{OF}_{2}\) (b) \(\mathrm{KrF}_{4}\) (c) \(\mathrm{CF}_{4}\) (d) \(\mathrm{GeCl}_{4}\)

Indicate which, if any, of the listed bond angles occur in the following species: (a) \(\mathrm{TeF}_{6}\) (1) \(90^{\circ}\) (b) \(\mathrm{SbCl}_{5}\) (2) \(109.5^{\circ}\) (c) \(\mathrm{ICl}_{\overline{4}}^{-}\) (3) \(120^{\circ}\) (d) \(\operatorname{InBr}_{3}\)

Give the molecular class and steric number and name the geometry that describes the shape of each of the following molecules: (a) \(\mathrm{SF}_{4}\) (b) \(\mathrm{SF}_{6}\) (c) \(\mathrm{XeOF}_{4}\) (d) IOF
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